Question

@mathematics
find the absolute max and absolute min values
f(t)=t(sqrt of 4-t^2) , [-1,2] @mathematics
find the absolute max and absolute min values
f(t)=t(sqrt of 4-t^2) , [-1,2] @Mathematics

Answer 1

You're in luck, i just had one of these questions on my homework :D.. Ok first thing to do is find the derivative and set it equal to zero.

Answer 2

do i do the chain rule?

Answer 3

Chain rule and product rule.

Answer 4

\[f(t)=t\sqrt{4-t^2}\]
\[f'(t)=\sqrt{4-t^2}+\frac{-t}{\sqrt{4-t^2}}\]

Answer 5

ok and then what?\

Answer 6

Once you find the derivative set it equal to 0.. So for the derivative i got \[\frac{4 - 2t^2}{\sqrt{4-t^2}}\]. So set the numerator equal to 0 and solve for t

Answer 7

subtract. you need to multiply first term top and bottom by
\[\sqrt{4-t^2}\] so you get
what tyler said

Answer 8

you can ignore the denominator and just set the numerator = 0 and solve

Answer 9

thank you both :)

Answer 10

well maybe not. the denominator is 0 at 0 so that is a critical point as well.

Answer 11

yeah i was going to say you have to check where the derivative doesnt exist.

Answer 12

so you have to check
\[f(-1),f(0),f(\sqrt{2}), f(2)\] the biggest is the biggest and the smallest is the smallest

Answer 13

So your critical points are \[\pm \sqrt{2}\] and 0

Answer 14

Your interval is [-1,2] and since \[- \sqrt{2}\] isnt in your interval you can forget about that number.